The scores on a statewide math exam were normally distributed with $\mu = 72.62$ and $\sigma = 6$. Nadia earned a n $89$ on the exam. Nadia's exam grade was higher than what fraction of test-takers? Use the cumulative z-table provided below. z.00.01.02.03.04.05.06.07.08.09 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974 2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981 2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986
Explanation: A cumulative z-table shows the probability that a standard normal variable will be less than a certain value (z) In order to use the z-table, we first need to determine the z-score of Nadia's exam grade. Recall that we can calculate her z-score by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}} = \dfrac{89 - {72.62}}{{6}} = 2.73} $ Look up $2.73$ on the z-table. This value, $0.9968$ , represents the portion of the population that scored lower than $89$ on the exam. Nadia scored higher than $99.68\%$ of the test-takers on the math exam.